# Fundamental frequency of a wire wheel spoke

## Homework Statement

The spoke of a wire wheel is 9.5 cm long, 3.5 mm in diameter, and under tension of 2100 N. The wire is made of steel of density 7860 kg/m3. When struck with a metal tool at its center, the spoke rings at its fundamental frequency. What is that frequency?

## Homework Equations

λn.fn=v and velocity:v=sqrt(T/μ) , lemda λn=2L/n and L= length of string
fn=n/2L . sqrt(T/μ)
fundamental≡ the first mode ⇒ n=1
μ is the linear mass and must be in kg/m so we must multiply our density by the area which is π.r^2

## The Attempt at a Solution

μ=7860*π*(3.5*10^-3)^2=0.302 kg/m
f=1/(2*9.5*10^-2) . sqrt(2100/0.302)
f= 438.8 Hz
but the answer is incorrect why??

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gneill
Mentor

μ=7860*π*(3.5*10^-3)^2=0.302 kg/m
Were you given the radius of the spoke?

• Any Help
Were you given the radius of the spoke?
Ah okay. Thanks very much. I used the diameter as a radius wrongly. :)
Now it will be 877Hz which is the correct answer